Question

#Quality Question
@ Integrals

Find Value of
$\huge \bf \int cos (log_{e}x)$
​

Answer 1

Step-by-step explanation:

Let I=∫cos(lnx)dx

I=cos(lnx)⋅x+∫sin(lnx)dx

cos(lnx)x+[sin(lnx)⋅x−∫cos(lnx)dx]

I= x/2sin(lnx)+cos(lnx)]+C

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle \rm \int cos( log_{e}x) \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle \rm \int \: cos( log_{e}x) \: dx$

We use here method of Substitution,

$\red{\rm :\longmapsto\: \: Put \: log_{e}(x) = y}$

$\red{\rm :\longmapsto\:x = {e}^{y}}$

$\red{\rm :\longmapsto\:dx = {e}^{y}dy}$

So, given integral can be rewritten as,

$\rm :\longmapsto\:I = \displaystyle \rm \int cosy \: {e}^{y}dy$

Now, using by parts, we get

$\rm :\longmapsto\:I = \displaystyle \: cosy \rm \int{e}^{y}dy - \displaystyle \rm \int \bigg(\dfrac{d}{dy}cosy\displaystyle \rm \int {e}^{y}dy \bigg)dy$

$\boxed{ \because \rm \: \displaystyle \rm \int uv \: dx = u\displaystyle \rm \int vdx - \displaystyle \rm \int \bigg(\dfrac{d}{dx}u\displaystyle \rm \int vdx \bigg) dx}$

$\rm :\longmapsto\:I = cosy \: {e}^{y} + \displaystyle \rm \int siny \: {e}^{y}dy$

Now, again using By Parts, we get

$\rm :\longmapsto\:I = cosy \: {e}^{y} +siny \displaystyle \rm \int {e}^{y}dy - \displaystyle \rm \int \bigg(\dfrac{d}{dy}siny\displaystyle \rm \int {e}^{y}dy\bigg) dy$

$\boxed{ \because \rm \: \displaystyle \rm \int uv \: dx = u\displaystyle \rm \int vdx - \displaystyle \rm \int \bigg(\dfrac{d}{dx}u\displaystyle \rm \int vdx \bigg) dx}$

$\rm :\longmapsto\:I = {e}^{y}cosy + {e}^{y}siny - \displaystyle \rm \int cosy \: {e}^{y}dy$

$\rm :\longmapsto\:I = {e}^{y}cosy + {e}^{y}siny - I$

$\rm :\longmapsto\:2I = {e}^{y}cosy + {e}^{y}siny$

$\rm :\longmapsto\:I = \dfrac{1}{2}\bigg({e}^{y}cosy + {e}^{y}siny\bigg) + c$

$\rm :\longmapsto\:I = \dfrac{{e}^{y}}{2}\bigg(cosy + siny\bigg) + c$

On substituting back the values, we get

$\rm :\longmapsto\:I = \dfrac{x}{2}\bigg(cos( log_{e}x) + sin( log_{e}x) \bigg) + c$

Basic Concept Used :-

Integration by Parts See the rule:

$\boxed{ \rm \: \displaystyle \rm \int uv \: dx = u\displaystyle \rm \int vdx - \displaystyle \rm \int \bigg(\dfrac{d}{dx}u\displaystyle \rm \int vdx \bigg) dx}$

Where,

u is the function u(x)

v is the function v(x)

u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

I - Inverse trigonometric functions

L -Logarithmic functions

A - Arithmetic and Algebraic functions

T - Trigonometric functions

E- Exponential functions

The alphabet which comes first is choosen as u and other as v.